I'm reading Learn You a Haskell (loving it so far) and it teaches how to implement elem in terms of foldl, using a lambda. The lambda solution seemed a bit ugly to me so I tried to think of alternative implementations (all using foldl):
import qualified Data.Set as Set
import qualified Data.List as List
-- LYAH implementation
elem1 :: (Eq a) => a -> [a] -> Bool
y `elem1` ys =
foldl (\acc x -> if x == y then True else acc) False ys
-- When I thought about stripping duplicates from a list
-- the first thing that came to my mind was the mathematical set
elem2 :: (Eq a) => a -> [a] -> Bool
y `elem2` ys =
head $ Set.toList $ Set.fromList $ filter (==True) $ map (==y) ys
-- Then I discovered `nub` which seems to be highly optimized:
elem3 :: (Eq a) => a -> [a] -> Bool
y `elem3` ys =
head $ List.nub $ filter (==True) $ map (==y) ys
I loaded these functions in GHCi and did :set +s and then evaluated a small benchmark:
3 `elem1` [1..1000000] -- => (0.24 secs, 160,075,192 bytes)
3 `elem2` [1..1000000] -- => (0.51 secs, 168,078,424 bytes)
3 `elem3` [1..1000000] -- => (0.01 secs, 77,272 bytes)
I then tried to do the same on a (much) bigger list:
3 `elem3` [1..10000000000000000000000000000000000000000000000000000000000000000000000000]
elem1 and elem2 took a very long time, while elem3 was instantaneous (almost identical to the first benchmark).
I think this is because GHC knows that 3 is a member of [1..1000000], and the big number I used in the second benchmark is bigger than 1000000, hence 3 is also a member of [1..bigNumber] and GHC doesn't have to compute the expression at all.
But how is it able to automatically cache (or memoize, a term that Land of Lisp taught me) elem3 but not the two other ones?